A chord of a circle of radius 10 cm subtends a right angleat its centre. The length of the chord (in cm) is

5 $\sqrt{2}$

10 $\sqrt{2}$

$\frac{5}{\sqrt{2}}$

10 $\sqrt{3}$

Solution

10 $\sqrt{2}$

$Given : ∠ AOB is given as 90 ° ∆ AOB is an isosceles trianglesince OA = OB Therefore ∠ OAB = ∠ OBA = 45 ° Thus ∠ AOP = 45 ° and ∠ BOP = 45 ° Hence ∆ AOP and ∆ BOP also are isosceles triangles Thus let AP = PB = OP = x Using pythagoras theorem x^{2} + x^{2} = 10^{2} Thus 2 x^{2} = 100 x = 5 \sqrt{2} Hence length of chord AB = 2 x = 10 \sqrt{2}$

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Circles

A chord of a circle of radius 10 cm subtends a right angleat its centre. The length of the chord (in cm) is

5 $\sqrt{2}$

10 $\sqrt{2}$

$\frac{5}{\sqrt{2}}$

10 $\sqrt{3}$

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A chord of a circle of radius 10 cm subtends a right angleat its centre. The length of the chord (in cm) is 52 102 52 103

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Entrance Exams Preparation

A chord of a circle of radius 10 cm subtends a right angleat its centre. The length of the chord (in cm) is

5 $\sqrt{2}$

10 $\sqrt{2}$

$\frac{5}{\sqrt{2}}$

10 $\sqrt{3}$